Solution Verification: Prove that $S \setminus C$ is finite or countable.

Question

I'm currently reading Berkeley Problems in Mathematics. I'm having troubles understanding the first problem/solution on metric spaces.

Problem:

Let $S$ be a subset of $\mathbb{R}$. Let $C$ be the set of points $x$ in $\mathbb{R}$ with the property that $S\cap (x-\delta,x+\delta)$ is uncountable for every $\delta \gt 0$. Prove that $S \setminus C$ is finite or countable.

Solution:

Suppose $x$ is a point of $S \setminus C$. Then there is an open interval containing $x$ whose intersection with $S$ is finite or countable and which thus is disjoint from $C$. By the density of $\mathbb{Q}$ in $\mathbb{R}$, there is such an interval with rational endpoints. There are countably many open intervals in $\mathbb{R}$ with rational endpoints. Hence $S \setminus C$ is a finite or countable union of finite or countable sets. Therefore, $S \setminus C$ is finite or countable.

If I understand correctly, $C$ is not the set of points, but the set of all points with the stated property ( $ S \cap (x-\delta,x+\delta)$ uncountable for every $\delta \gt 0$).

Some other weird thing in the solution is the fact that they say that the intersection of $S$ with the open interval containing one point $x$ can be finite, how can it be finite when it corresponds to a subset of $\mathbb{R}$? is it because the only way it can be finite is when the subset $S$ consist of a set of points? because if $S$ is an interval, if you have an non-empty intersection, it's going to be either one single point or some other interval (which is uncountable), is there other case?.

Again, with all this, the solution seems to me a big vague, but it might be that I just don't have the right background (yet).

If somebody have another way to express the solution, that might also help.

Answer 1

By definition $x \in C$ if and only if

$$ S \cap (x-\delta , x+ \delta )$$ is uncountable for all $\delta >0$

All you need for a point $x$ not be in the set $C$ is for you to find a value $\delta $ such that $S \cap (x- \delta,x+\delta )$ is not uncoutable (or in other words countable or finite).

Some other weird thing in the solution is the fact that they say that the intersection of S with the open interval containing one point x can be finite, how can it be finite when it corresponds to a subset of R ?

At the start of the proof it says that $x \in S \setminus C$. This means that $x \not \in C$ so as I said above this means that by definition there exists a value for $\delta $ such that the set $S \cap (x-\delta , x+ \delta )$ is finite or countable. Moreover it is not a problem for subsets of $\mathbb R $ to be finite or countable. There are plenty such subsets (for example the set $\{1,2,3\}$ which is finite, or the set of rational numbers which is countable. )